Giải thử:
\(\sqrt{1+2+3+...+\left(n-1\right)+n+\left(n-1\right)+...+3+2+1}\)
\(=\sqrt{2.\left[1+2+3+...+\left(n-1\right)\right]+n}\)
\(=\sqrt{2.\frac{\left[\left(n-1\right)+1\right].\left(n-1\right)}{2}+n}\)
\(=\sqrt{n.\left(n-1\right)+n}\)
\(=\sqrt{n\left(n-1+1\right)}\)
\(=\sqrt{n^2}\)
\(=n\left(v\text{ì}n>0\right)\)
đpcm
kudo shinichi đúng r,mình nêu cách của mình:
\(VT=\sqrt{2\left[1+2+3+...+\left(n+1\right)+n\right]-n}\)
\(=\sqrt{2.\frac{\left(n+1\right)n}{2}-n}=\sqrt{\left(n+1\right)n-n}\)
\(=\sqrt{n^2+n-n}=\sqrt{n^2}=n^{\left(đpcm\right)}\)